Notes on string theory #5: Modifying the Polyakov action (cosmological constant and Ricci scalar)

We come to the conclusion of Section 1.2 in Polchinski’s textbook. In this note we’ll discuss how there are two possible modifications that we can make to the Polyakov action (see last note) that preserve Poincaré invariance. The first is a cosmological constant term on the worldsheet. The second modification involves the scalar curvature ${R}$ of the 2-dimensional worldsheet geometry. This particular modification is given in eqn.(1.2.31) in the textbook.

To arrive at these two possible modifications we will take a slight detour, beginning with a discussion of General Relativity in M-dimensions, which also provides an alternative route to the Polyakov action. Our main focus at the end will be to derive an important equation, namely eqn. (1.2.32) in the textbook, which describes the Weyl rescaling of ${R}$ . (This note was one of the messiest in my folder, as it goes from General Relativity to the point particle and string actions. I’ve quickly tidied it a bit as I tried to remember what I was thinking a few years ago, so hopefully it is clear).

2. General Relativity in M-dimensions

We start by thinking about General Relativity in M-dimensions, with the inclusion of scalar fields and the cosmological constant (note that the following section is based on a lecture by Freddy Cachazo that I listened to a couple of years ago but which I can no longer locate).

Start with a simple scalar field theory with the action given by

$\displaystyle S[g, \phi] = \int d^{d}x \sqrt{-g} (R - \frac{1}{2} g^{mn} \partial_{m} \phi \partial_{n} \phi - \Lambda), \ \ \ \ \ (1)$

with ${R}$ the Ricci scalar and ${\Lambda}$ the cosmological constant. From GR, we know that this action is of course diffeomorphism invariant. It also has global symmetry, such that ${\phi \rightarrow \phi + a}$ .

What we want to do is replace ${\partial_{m} \phi \partial_{n} \phi}$ with a linear combination of the products of many scalar fields. So we’re going to take the conventional route of employing ${i}$ and ${j}$ indices, with the inclusion of some matrix ${M_{ij}}$ in the action. Here, ${M_{ij}}$ is not a function of ${\phi}$ – it is just a matrix of numbers. Taking the described steps gives an action of the form

$\displaystyle S[g, \phi] = \int d^{d}x \sqrt{-g} (R - \frac{1}{2} g^{mn} \partial_{m} \phi^{i} \partial_{n} \phi^{j} M_{ij} - \Lambda). \ \ \ \ \ (2)$

Several comments are in order: first, the ${i}$ and ${j}$ indices represent the internal space of our theory where ${\phi}$ can take values. Secondly, the global symmetry that we had before remains such that ${\phi^i \rightarrow \phi^i + a^i}$ . But there is also another global symmetry that we can write, because if we transform the scalars by any matrix ${\phi^i \rightarrow \Lambda_{j}^{i} \phi^i}$ such that ${\Lambda^{T} M \Lambda = M}$ , we have another global symmetry for the action.

Furthermore, if we choose a particular ${M}$ such that it equals the identity ${M = \mathbb{I}_{D\times D}}$ with ${i = 1, ... , D}$ , the global symmetry group can be written as ${SO(D)}$ . If, instead, we choose the matrix

$\displaystyle M = \begin{pmatrix} -1 & 0 \\ 0 & \mathbb{I}_{(D-1)+(D-1)} \end{pmatrix}, \ \ \ \ \ (3)$

we can write the global symmetry group explicitly as ${SO(1, D-1)}$ . In QFT, this choice for ${M}$ is often avoided due to the desire for positive energy. That is because if this symmetry group is used, one of the fields will have the incorrect sign for the kinetic term. But let’s not worry about that. What is nice is that if we combine ${SO(1, D-1)}$ with the previous expression for the global symmetry ${\phi^i \rightarrow \Lambda^i + a^i}$ , we obtain something that looks like a translation. So the result is the combination of translations plus the Lorentz group together to form the Poincaré group.

Even more interestingly, if we maintain this choice of ${M}$ and we choose ${\phi^i \rightarrow \phi^i + a^i}$ , what we’re looking at for the internal space is something that particularly resembles Minkowski space. Indeed, returning to the action (2), when we vary the metric the equations of motion can be found to take the form

$\displaystyle \frac{\delta S}{\delta g_{mn}} = 0 \rightarrow \mathcal{R}_{m} - \frac{1}{2} g_{mn} \mathcal{R} + \Lambda g_{mn} = T_{mn}. \ \ \ \ \ (4)$

This is our first important equation. Notice that ${\mathcal{R}_{m} - \frac{1}{2} g_{mn} \mathcal{R}}$ is the Einstein tensor. On the right-hand side, ${T_{mn}}$ is the energy-momentum tensor of the matter theory, which we discussed in the string setting in the last note. One can quickly verify that the energy-momentum tensor can be written as

$\displaystyle T_{mn} = M_{ij} (\partial_{m} \phi^{i} \partial_{n} \phi^{j} - \frac{1}{2} g_{mn} g^{rs} \partial_{r} \phi^{i} \partial_{s} \phi^{j}). \ \ \ \ \ (5)$

What we will want to do is return to these two equations. But first, let’s lay some more ground work.

2.1. ${M=1}$

We want to think about different cases of ${M}$ for the theory we’ve constructed. Consider the case ${M = 1}$ – if there is only 1-dimension, as illustrated in a past post on point particle theory – then the only dimension we have is time. So, ${X^0 \rightarrow \tau}$ . We can also establish that the metric ${M_{ij} \rightarrow \eta_{\mu \nu}}$ , as we’ll be working in a sort of Minkowski configuration. So we will also replace the previous notation with more appropriate symbols for the given context, as our scalar fields no longer carry ${i}$ indices: namely, ${\phi_{(\tau)}^{i} \rightarrow X_{(\tau)}^{\mu}}$ .

As for the metric ${g_{mn}}$ , in the case of ${M=1}$ it only has one degree of freedom, and the convention employed (used for convenience) is simply to define ${g = -e^{2}}$ . As there is only a time component, the idea then is to define ${e}$ as positive with a minus sign in front. This ${e}$ term is of course also a field – in fact, one can think of it as an auxiliary field – and it should be clear that ${X}$ is also a field. We also have that in 1-dimension ${R \rightarrow 0}$ , which is to say there is no Riemann tensor. All of this is made quite explicit in the new action

$\displaystyle S[e, X] = - \frac{1}{2} \int d\tau (\frac{1}{-e^2} \dot{X}^{\mu} \dot{X}^{\nu} \eta_{\mu \nu} - 2e \Lambda). \ \ \ \ \ (6)$

where we employ the familiar short-hand ${\partial_{\tau} X^{\mu} \equiv \dot{X}^{\mu}}$ .

Next, we want to calculate the equations of motion for the field ${e}$ , with the hope that something interesting might ‘pop out’. We could use the energy-momentum tensor instead; however, the above action is quite simple. So, we can compute the equations of motion directly giving

$\displaystyle \frac{\delta S}{\delta e} = 0 \rightarrow (-e^{-2} \dot{X}^{\mu} \dot{X}^{\nu} \eta_{\mu \nu} - 2 \Lambda) = 0. \ \ \ \ \ (7)$

The key observation is that ${e}$ can be integrated out. This means we can try to find the equation of motion, solving for ${e}$ , and then substituting this value back into the action. Setting ${\dot{X}^{\mu} \dot{X}^{\nu} \eta_{\mu \nu}}$ to be ${\dot{X}^2}$ for simplification purposes we find

$\displaystyle -e^{-2} = \frac{2 \Lambda}{\dot{X}^2}$

$\displaystyle e^{2} = -\frac{\dot{X}^2}{2 \Lambda}$

$\displaystyle \implies e = -\frac{\sqrt{\dot{X}^2}}{\sqrt{2 \Lambda}}. \ \ \ \ \ (8)$

Thus, the effective action with just ${X}$ remaining has the form

$\displaystyle S_{eff}[X] = \frac{1}{2} \int d\tau \ - \frac{\sqrt{2 \Lambda}}{ \sqrt{\dot{X}^2}}\dot{X}^2 - 2 \frac{\sqrt{\dot{X}^2}}{\sqrt{2 \Lambda}} \Lambda$

$\displaystyle = -\sqrt{2 \Lambda} \int d\tau \sqrt{\dot{-X}^2}. \ \ \ \ \ (9)$

This is beginning to look very familiar, if we compare with past notes. Putting this result to one side for the moment, if we consider the non-relativistic limit notice we arrive at an expression for ${\sqrt{2 \Lambda}}$ that simplifies the above a bit more. Working with a metric ${\eta_{\mu \nu}}$ that is mostly plus we can write,

$\displaystyle \dot{X}^2 = -(\frac{d\tau}{d\tau})^2 + (\frac{d\vec{x}}{d \tau})^2. \ \ \ \ \ (10)$

Therefore, we can rewrite (9) as

$\displaystyle S = - \sqrt{2 \Lambda} \int d\tau \sqrt{1- (\frac{d\vec{x}}{d \tau})^2}. \ \ \ \ \ (11)$

Approximating the term under the square root, we have seen in a previous note that we get ${\sqrt{1- (\frac{d\vec{x}}{d \tau})^2} \sim 1 - \frac{1}{2} (\frac{d\vec{x}}{d \tau})^2}$ . It follows that

$\displaystyle S = \int d\tau (-\sqrt{2 \Lambda} + \frac{\sqrt{2 \Lambda}}{2} (\frac{d\vec{x}}{d \tau})^2 + ...). \ \ \ \ \ (12)$

Ignoring the linear term, we see that ${\frac{\sqrt{2 \Lambda}}{2} (\frac{d\vec{x}}{d \tau})^2}$ resembles the action of a point particle – namely, the kinetic term. If, in other words, we interpret ${\sqrt{2 \Lambda} = m}$ and plug this back into the action, then clearly we have a theory of a point particle propagating in flat spacetime as constructed in a past note:

$\displaystyle S_{eff}[x] = -m \int d\tau \sqrt{-\dot{X}^2}. \ \ \ \ \ (13)$

Replacing the metric ${\eta_{\mu \nu}}$ with a general metric ${G_{\mu \nu}(X)}$ that depends on the field ${X}$ , we have

$\displaystyle S[X] = -m \int d\tau \sqrt{- \dot{X}^{\mu} \dot{X}^{\nu} G_{\mu \nu}}. \ \ \ \ \ (14)$

Now we have an action for a particle moving in a curved spacetime. It is a nice result, given it connects with past entries and considerations.

2.2. ${M=2}$

Let’s now attempt to translate the general field theory to the case of ${M=2}$ . We will first update our notation because we now have the coordinates ${(X^0, X^1) \rightarrow (\tau, \sigma)}$ , which should also look familiar from past discussions. The scalar fields are still ${\phi^i \rightarrow X^{\mu}(\tau, \sigma)}$ as we will also set ${M_{ij} \rightarrow \eta_{\mu \nu}}$ . In this space, we will also write ${g_{mn} = \gamma_{mn}}$ .

Even though we can more or less translate everything we’ve already defined to the ${M=2}$ case, we do have one problem. We need to update our equations of motion, and this also entails that we define the Riemann tensor in the updated context. This requires that we recall some basic facts about the Riemann tensor in ${M}$ -dimensions, including its symmetry relations and components, where it can be checked that the number of degrees of freedom is given by

$\displaystyle R^{q}_{mnp} = \frac{1}{12} M^2 (M^2 - 1). \ \ \ \ \ (15)$

If ${M=2}$ , then

$\displaystyle R^{q}_{mnp} = \frac{1}{12} (2)^2 (2^2 - 1) = 1. \ \ \ \ \ (16)$

This means that we only have one parameter or degree of freedom.

We can write the correct symmetry properties for the metric as follows

$\displaystyle R_{mnpq} = \alpha (g_{mp} g_{qn} - g_{mq}g_{pn}), \ \ \ \ \ (17)$

where ${\alpha}$ is the degree of freedom or, in other words, the constant of proportionality. We need to find this constant of proportionality if we’re to progress. The key is this: if ${\alpha}$ is proportional to the Ricci scalar – the only other scalar in our theory – we can try to produce the Ricci scalar on the left-hand side and then solve for ${\alpha}$ . The first step is to take the Ricci tensor, contract indices, and then simplify:

$\displaystyle g^{mp} R_{mnpq} = R_{nq}$

$\displaystyle \implies R_{nq} = \alpha (2 g_{qn} - g_{qn}) = \alpha g_{qn}$

$\displaystyle \implies R = g^{nq} R_{nq} = 2\alpha$

$\displaystyle \alpha = \frac{R}{2}. \ \ \ \ \ (18)$

We can now substitute this result back into the second line above giving

$\displaystyle R_{nq} = \frac{R}{2} g_{qn} \implies R_{nq} - \frac{R}{2} g_{qn} = 0. \ \ \ \ \ (19)$

With this result, we have to use a bit of logic. Let’s return all the way back to the equation (4). What we see in the fact that the Einstein tensor vanishes is that from the equations of motion the following must be true

$\displaystyle \Lambda g_{mn} = T_{mn}. \ \ \ \ \ (20)$

The pressing question is what do we make of the energy-momentum tensor? We already know what it is from the discussion above, so let’s translate it for ${M=2}$ case:

$\displaystyle T_{mn} = \eta_{\mu \nu} (\partial_{m} X^{\mu} \partial_{n} X^{\nu} - \frac{1}{2} \gamma_{mn} \gamma^{pq} \partial_{p} X^{\mu} \partial_{q} X^{\mu}). \ \ \ \ \ (21)$

Taking the trace we indeed find as we did in the last note,

$\displaystyle T_{m}^{m} = \gamma^{mn}T_{mn} = 0. \ \ \ \ \ (22)$

But if the trace of the energy-momentum tensor is zero, it follows

$\displaystyle \Lambda g_{mn} = T_{mn} \rightarrow \Lambda = 0. \ \ \ \ \ (23)$

Thus, the cosmological constant vanishes. So for the equations of motion to be consistent ${\Lambda = 0}$ .

If we now travel back to the master action (2), the lingering question is what happens to gravity such that the Einstein tensor equals zero? The answer is that, when we translate the old formula using updated notation in the context ${M=2}$ , we have the object

$\displaystyle S = \int d\tau d\sigma \sqrt{-\gamma} R, \ \ \ \ \ (24)$

because as we have found so far (18) the Ricci scalar is not zero. And yet, this action (24) doesn’t give equations of motion that are interesting. Why? In 2-dimensions, this integrand is just a constant. The number it spits out describes the topology of the space. So, more formally, if we go to Euclidean space such that ${\tau \rightarrow i\tau}$ and we have some closed surface ${M}$ , then the total object ${\int_{M} d\tau d\sigma \sqrt{-\gamma} R}$ produces the Euler characteristic of this particular 2-dimensional space

$\displaystyle S = \int_{M} d\tau d\sigma \sqrt{-\gamma} \ R = \chi(M). \ \ \ \ \ (25)$

The important point is that, as this action is just a number, we don’t really see it in the 2-dimensional action unless we make it explicit. What we therefore end up with, after all that, is a 2-dimensional action for our metric and fields (where we set ${\tau = \sigma^{0}, \sigma=\sigma^{1})}$ of the form

$\displaystyle S_{p} = \int d^2 \sigma \sqrt{-\gamma} (- \frac{1}{2} h^{mn} \partial_{m} X^{\mu} \partial_{n} X^{\nu} \eta_{\mu \nu}), \ \ \ \ \ (26)$

which of course is just the Polyakov action (without the inclusion of the slope parameter). But if we do include (24)and make it explicit in the action we have

$\displaystyle S_P = - \int_M d\tau d\sigma (-\gamma)^{1/2} (\frac{1}{4\pi \alpha^{\prime}} \gamma^{ab}\partial_a X^{\mu}\partial_b X_{\mu} + \frac{\lambda}{4\pi} R), \ \ \ \ \ (27)$

which is of the form of eqn. (1.2.33) in Polchinski.

To summarise, we have seen that although it is possible to add a cosmological constant term (here written in the string worldsheet theory) ${\lambda \int d^2 \sigma \sqrt{-\gamma}}$ , this inclusion is ultimately not allowed by the equations of motion. Another, more direct way to see this is to take the string sigma model action and add this cosmological constant term, whereby consistency of the equations of motion requires ${\lambda = 0}$ just as we saw. (Note, however, that the sigma model action of a p-brane for ${p \neq 1}$ requires a cosmological constant term!). The second modification to the Polyakov action involves the scalar curvature ${R^{(2)}(\gamma)}$ of the 2-dimensional worldsheet geometry. As we just saw, this is simply a constant. If we choose to include it, then it is proposed that the most general diffeomormphism, Weyl, and Poincare invariant action is of the form (27).

2.3. Weyl rescaling

As Polchinski notes on p. 15: “symmetry, not simplicity, is the key idea”. If (27) is allowed, it is so by the symmetries. We already know that this generalisation of the Polyakov action is invariant under Poincaré and diffeomorphisms. The last symmetry of the string action that must be maintained is invariance under Weyl rescaling.

Proposition 1 As given in eqn. (1.2.32) in Polchinski, under local Weyl rescaling we have

$\displaystyle (-\gamma^{\prime})^{1/2} R^{\prime} = ((-\gamma)^{1/2}R - 2 \nabla^2 \omega). \ \ \ \ \ (28)$

Let’s think about this equation. On the left-hand side, we have the transformed Ricci scalar ${R^{\prime}}$ . On the right-hand side, we have the untransformed ${R}$ and in the last term there is a second-order covariant derivative of the Weyl factor ${\omega}$ . As this is an important equation, it is worth deriving. There are a few ways to go about this task. The long and tedious way is to first calculate the Ricci scalar ${R^{\prime}}$ , which is calculated from the Christoffel symbols and their derivatives. As (28) describes a transformation under Weyl rescaling, we should then also substitute ${\gamma^{\prime ab} = e^{-2\omega}\gamma^{ab}}$ where appropriate.

Proof: We start with some standard definitions from differential geometry, beginning with the definition of the Riemann curvature tensor in terms of the metric ${\gamma}$

$\displaystyle R^d_{abc} = \Gamma^d_{ac, b} - \Gamma^d_{ab, c} + \Gamma^{d}_{be}\Gamma^e_{ac} - \Gamma^d_{ce}\Gamma^e_{ab}, \ \ \ \ \ (29)$

with the connection defined as

$\displaystyle \Gamma^{c}_{ab} = \frac{1}{2}\gamma^{cd}(\gamma_{bd,a} + \gamma_{ad,b} - \gamma_{ab,d}). \ \ \ \ \ (30)$

It clearly follows that for the Ricci curvature we have

$\displaystyle R = g^{ab}R_{ab} = g^{ab}R^{d}_{acb} = g^{ab}(\Gamma^d_{ac, d} - \Gamma^d_{ab, c} + \Gamma^{d}_{be}\Gamma^e_{ac} - \Gamma^d_{ce}\Gamma^e_{ab}), \ \ \ \ \ (31)$

where ${\Gamma^d_{ac}}$ are Christoffel symbols of the metric and ${\Gamma^d_{ac, d}}$ is the partial derivative of ${\Gamma^d_{ac}}$ .

Now, consider a Weyl transformation of the metric ${\gamma^{\prime}_{ab} = e^{2\omega(\sigma)}\gamma_{ab}}$ . Except in the present case indices are raised, so we pick up a minus sign ${\gamma^{\prime ab} = e^{- 2\omega(\sigma)}\gamma^{ab}}$ . Substituting in (30) for the transformed metric we find

$\displaystyle \Gamma^{\prime c}_{ab} = \frac{1}{2}\gamma^{\prime cd}(\gamma^{\prime}_{bd,a} + \gamma^{\prime}_{ad,b} - \gamma^{\prime}_{ab,d})$

$\displaystyle = \frac{1}{2} e^{- 2\omega(\sigma)} \gamma^{cd} [e^{2\omega}(\gamma_{bd,a}) + e^{2\omega}(\gamma_{ad,b}) - e^{2\omega}(\gamma_{ab,d})]. \ \ \ \ \ (32)$

We can pull the partial derivatives outside the brackets, e.g., ${e^{2\omega}(\gamma_{bd,a}) = \partial_a (e^{2\omega}\gamma_{bd})}$ . Then, using the product and chain rules, we compute each term to find

$\displaystyle \Gamma^{\prime c}_{ab} = \frac{1}{2} \gamma^{cd} (\partial_a \gamma_{bd} + \partial_b \gamma_{ad} - \partial_d \gamma_{ab}) + \gamma^{cd}(\gamma_{bd}\partial_a \omega + \gamma_{ad}\partial_b \omega - \gamma_{ab}\partial_d \omega)$

$\displaystyle =\Gamma^{c}_{ab} + \Delta^{c}_{ab}, \ \ \ \ \ (33)$

in which ${\Delta^{c}_{ab}= \gamma^{cd}(\gamma_{bd}\partial_a \omega + \gamma_{ad}\partial_b \omega - \gamma_{ab}\partial_d \omega)}$ .

Now, a quick way to complete proof is to understand that by general covariance the result needs to be of the form

$\displaystyle -\gamma^{1/2} (R+ a\nabla \omega \cdot \nabla \omega + b\nabla 2\omega). \ \ \ \ \ (34)$

To see this, we know that ${R}$ is a scalar under diffeomorphisms, and so we can write everything in terms of covariant derivatives. Furthermore, as the Ricci scalar contains connections which contain derivatives of the Weyl factor ${\omega}$ , these are the terms of interest. What we therefore want to do is find the coefficients ${a}$ and ${b}$ .

It’s easiest to start by working out the ${b}$ coefficient by first noting that it’s attached to a second-order derivative. We can match this in ${R^{\prime}}$ by writing out the transformed Ricci scalar in terms of Christoffel symbols and their derivatives

$\displaystyle R^{\prime d}_{acb} = \partial_d (\Gamma^d_{ca} + \Delta^d_{ca}) - \partial_b (\Gamma^d_{ba} + \Delta^d_{ba}) + (\Gamma^d_{be} + \Delta^d_{be})(\Gamma^e_{ca} + \Delta^e_{ca}) - (\Gamma^d_{ce} + \Delta^d_{ce})(\Gamma^e_{ba} + \Delta^e_{ba}), \ \ \ \ \ (35)$

where only the pieces with partial derivatives in front, i.e. ${\partial_d \Delta^d_{ca}}$ terms, contain second order derivatives. Hence, we consider

$\displaystyle R^{\prime} = \gamma^{\prime ab} R^{\prime d}_{acb} = \gamma^{\prime ab}[\partial_d (\Gamma^d_{ca} + \Delta^d_{ca}) - \partial_b (\Gamma^d_{ba} + \Delta^d_{ba}) + ...]$

$\displaystyle = e^{-2\omega} \gamma^{ab} (\partial_d \Delta^d_{ca} - \partial_b \Delta^d_{ba} + ...). \ \ \ \ \ (36)$

We now need to calculate each ${\gamma^{ab}\partial_d \Delta^d_{ca}}$ piece, which should give second order partial derivatives of ${\omega}$ . Working in general ${D}$ -dimensions, we first expand the covariant derivative with respect to the metric and then distribute the partial derivative. For the first term we find

$\displaystyle \gamma^{ab}\partial_d \nabla^d_{ca} = \gamma^{ab} \partial_d [\gamma^{de} (\gamma_{ae}\partial_b \omega + \gamma_{be} \partial_a \omega - \gamma_{ba} \partial_e \omega)]$

$\displaystyle = \gamma^{ab}(\delta^d_a \partial_d \partial_b \omega + \delta^d_b \partial_d \partial_a \omega - \gamma^{de}\gamma_{ba} \partial_d \partial_e \omega)$

$\displaystyle = \gamma^{db} \partial_d \partial_b \omega + \gamma^{ad}\partial_d \partial_a \omega - \delta^a_a \gamma^{de} \partial_d \partial_e \omega = (2-D)\partial^2 \omega, \ \ \ \ \ (37)$

where we picked up Kronecker delta functions and contracted and rearranged indices along the way.

A similar calculation for ${\gamma^{ab} \partial_b \Delta^d_{ba}}$ yields

$\displaystyle \gamma^{ab} \partial_b \Delta^d_{ba} = D\partial^2 \omega. \ \ \ \ \ (38)$

Returning to (28) we therefore find

$\displaystyle \gamma^{\prime 1/2 }R^{\prime} = e^{2\omega} \gamma^{1/2}e^{-2\omega}[(2 - D)\partial^2 \omega + D\partial^2 \omega] \ \ \ \ \ (39)$

which we can write as

$\displaystyle = \gamma^{1/2}[2 + 2D] \partial^2 \omega, \ \ \ \ \ (40)$

and so we find ${b = -2[D - 1]}$ . When ${D = 2}$ we find ${b = -2}$ . This matches the coefficient of the second term in (28).

Now, for the coefficient ${a}$ , we note that ${a\nabla \omega \cdot \nabla \omega = a \partial^b \omega \partial_b \omega}$ . Terms with this derivative structure in ${R^{\prime}}$ come from any pieces where there is a product of connections

$\displaystyle R^{\prime} = \gamma^{\prime ab} (\Gamma^{\prime d}_{be}\Gamma^{\prime e}_{ca} - \Gamma^{\prime d}_{ce}\Gamma^{\prime e}_{ba}) + ... \ \ \ \ \ (41)$

in which we again substitute ${\gamma^{\prime ab} = e^{-2\omega}\gamma^{ab}}$ obtaining

$\displaystyle e^{-2\omega} \gamma^{ab}(\Delta^d_{de}\Delta^{e}_{ca} - \Delta^{d}_{ce}\Delta^{e}_{ba}) + ... \ \ \ \ \ (42)$

We next calculate each product in the brackets. For the first term we find

$\displaystyle \Delta^d_{de}\Delta^{e}_{ca} = \gamma^{db}(\gamma_{bd}\partial_e \omega +\gamma_{be} \partial_d \omega - \gamma_{ed} \partial_b \omega)\gamma^{ef}(\gamma_{fc} \partial_a \omega + \gamma_{fa} \partial_c \omega - \gamma_{ac} \partial_f \omega),$

which upon rearranging terms, using the distribution property, and contracting gives

$\displaystyle = \gamma^{db}\gamma^{ef}[(\gamma_{bd}\gamma_{fc}\partial_e \omega \partial_a \omega + \gamma_{bd}\gamma_{fa} \partial_e \omega \partial_c \omega - \gamma_{bd}\gamma_{ac}\partial_e \omega \partial_f \omega) + (\gamma_{be}\gamma_{fc} \partial_d \omega \partial_a \omega + \gamma_{be}\gamma_{fa} \partial_d \omega \partial c \omega - \gamma_{be}\gamma_{ac} \partial_d \omega \partial_f \omega) + (- \gamma_{de}\gamma_{fc}\partial_b \omega \partial_a \omega - \gamma_{ed}\gamma_{fa}\partial_b \omega \partial_{c} \omega + \gamma_{ed}\gamma_{ac} \partial_b \omega \partial_f \omega)]$

$\displaystyle = \delta^d_d \partial_c \omega \partial_a \omega + \delta^d_d \partial_a \omega \partial_c \omega - \delta^d_d \gamma_{ac} \partial^f \omega \partial_f \omega + \partial_c \omega \partial_a \omega + \partial_a \omega \partial_c \omega - \gamma_{ac} \partial^f \omega \partial_f \omega - \partial_c \omega \partial_a \omega - \partial_a \omega \partial_c \omega + \gamma_{ac} \partial_f \omega \partial^f \omega$

$\displaystyle =2D \partial_c \omega \partial_a \omega - D \gamma_{ac} \partial^f \omega \partial_f \omega. \ \ \ \ \ (43)$

Similarly, we can compute ${\Delta^{d}_{ce}\Delta^{e}_{ba}}$ finding many cancellations to give

$\displaystyle \Delta^{d}_{ce}\Delta^{e}_{ba} = (D + 2) \partial_a \omega \partial_c \omega - 2 \gamma_{ac} \partial^f \omega \partial_f \omega. \ \ \ \ \ (44)$

Putting everything together we end up with

$\displaystyle e^{-2\omega} \gamma^{ac}(\Delta^d_{de}\Delta^{e}_{ca} - \Delta^{d}_{ce}\Delta^{e}_{ba}) = e^{-2\omega} \gamma^{ac}[(2D \partial_c \omega \partial_a \omega - D \gamma_{ac} \partial^f \omega \partial_f \omega) - ((D + 2) \partial_a \omega \partial_c \omega - 2 \gamma_{ac} \partial^f \omega \partial_f \omega)]$

$\displaystyle = e^{-2\omega} \gamma^{ac} [\partial_c \omega \partial_a \omega(D-2) - \gamma_{ac} \partial^f \partial_f (D-2)]$

$\displaystyle e^{-2\omega} (D-2)(1-D)\partial \omega \cdot \partial \omega. \ \ \ \ \ (45)$

Therefore, comparing with (28) we find that ${a = -(D-1)(D-2)}$ . Setting ${D=2}$ it follows ${a = 0}$ . Completing the proof, we have found (28). $\Box$

2.4. Invariance under Weyl rescaling

Polchinski notes that (24) is invariant under (28), as the variation is a total derivative due to the fact that

$\displaystyle (-\gamma)^{1/2} \nabla_a v^a = \partial_a ((-\gamma)^{1/2}v^a) \ \ \ \ \ (46)$

for any ${v^a}$ .

To see this, we already know that a Weyl transformation of the metric takes the form

$\displaystyle \gamma_{ab}^{\prime } \rightarrow e^{2\omega}\gamma_{ab}. \ \ \ \ \ (47)$

We have also proven that the the Ricci tensor transforms as

$\displaystyle R \rightarrow e^{-2\omega}(R - 2\nabla^2 \omega). \ \ \ \ \ (48)$

Putting everything together it follows

$\displaystyle \chi \rightarrow \frac{1}{4\pi} \int_M d\tau d\sigma \ \sqrt{-\gamma}\ (R - 2\nabla^2\omega). \ \ \ \ \ (49)$

To prove ${\chi}$ is invariant under local Weyl rescaling we are left to prove

$\displaystyle -\frac{1}{2\pi} \int_M d\tau d\sigma \sqrt{-\gamma} \nabla^2 \omega \ \ \ \ \ (50)$

is zero.

To show this, let us go to Euclidean space and start with the left-hand side of (46):

$\displaystyle (\gamma)^{1/2} \nabla_a v^a = (\gamma)^{1/2}(\partial^a v_a + \Gamma^{a}_{ab}v^b), \ \ \ \ \ (51)$

which, upon expanding the Christoffel symbol gives,

$\displaystyle (\gamma)^{1/2}[\partial^a v_a + \frac{1}{2}(\gamma^{ac}\partial_a \gamma_{cb} + \gamma^{ac}\partial_b \gamma_{ca} - \gamma^{ac}\partial_c \gamma_{ab})v^b]$

$\displaystyle = (\gamma)^{1/2}(\partial_a v^a + \frac{1}{2}\gamma^{ac}\partial_b \gamma_{ca}v^b. \ \ \ \ \ (52)$

For the right-hand side of (46), the calculation is similarly straightforward. Using the product rule and then also the identity ${\partial \sqrt{\gamma} = \frac{1}{2 \sqrt{\gamma}} \sqrt{\gamma} \gamma^{bc}\partial_a \gamma_{bc} v^a}$ we find

$\displaystyle \partial_a ((\gamma)^{1/2}v^a) = (\gamma)^{1/2}(\frac{1}{2}\gamma^{bc}\partial_a \gamma_{bc} v^a + \partial_a v^a). \ \ \ \ \ (53)$

Therefore, the left- and right-hand sides of (46) are equivalent. And so, back in Minkowski space and therefore reinstating the minus signs, we see that the variation (50) gives a total derivative

$\displaystyle \sqrt{-\gamma}(- 2\nabla^2 \omega) = \sqrt{-\gamma} \nabla_a (\nabla^a \omega) = \partial_a (\sqrt{-\gamma} \nabla^a \omega), \ \ \ \ \ (54)$

which is just Stokes theorem on the worldsheet without boundary. When a boundary is included, an additional surface term is required, which is the topic of Exercise 1.3. As this note is growing in length, we will not cover that here.

This completes Section 1.2. In the next note, we will begin to discuss the quantisation of the open string.

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